## Thinking Mathematically (6th Edition)

(a) The monthly payments are $\$450$The interest is$\$1200$ (b) The monthly payments are $\$293$The interest is$\$2580$ (c) With Loan A, the monthly payments are $\$157$more than the monthly payments with Loan B. With Loan B, the interest is$\$1380$ more than the interest with Loan A.
We can use this formula to calculate the payments for a loan: $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$ $PMT$ is the amount of the regular payment $P$ is the amount of the loan $r$ is the interest rate $n$ is the number of payments per year $t$ is the number of years (a) $PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}$ $PMT = \frac{(\$15,000)~(\frac{0.051}{12})}{[1-(1+\frac{0.051}{12})^{-(12)(3)}~]}PMT = \$450$ The monthly payments are $\$450$We can find the total amount paid.$\$450 \times 36 = \$16,200$The interest is the difference between the total amount paid and the amount of the loan.$I = \$16,200 - \$15,000 = \$1200$ The interest is $\$1200$(b)$PMT = \frac{P~(\frac{r}{n})}{[1-(1+\frac{r}{n})^{-nt}~]}PMT = \frac{(\$15,000)~(\frac{0.064}{12})}{[1-(1+\frac{0.064}{12})^{-(12)(5)}~]}$ $PMT = \$293$The monthly payments are$\$293$ We can find the total amount paid. $\$293 \times 60 = \$17,580$ The interest is the difference between the total amount paid and the amount of the loan. $I = \$17,580 - \$15,000 = \$2580$The interest is$\$2580$ (c) With Loan A, the monthly payments are $\$450$With Loan B, the monthly payments are$\$293$ With Loan A, the monthly payments are $\$157$more than the monthly payments with Loan B. With Loan A, the interest is$\$1200$ With Loan B, the interest is $\$2580$With Loan B, the interest is$\$1380$ more than the interest with Loan A.